You have found the following ages (in years) of all 4 turtles at your local zoo: $ 46,\enspace 67,\enspace 105,\enspace 57$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{46 + 67 + 105 + 57}{{4}} = {68.8\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $46$ years $-22.8$ years $519.84$ years $^2$ $67$ years $-1.8$ years $3.24$ years $^2$ $105$ years $36.2$ years $1310.44$ years $^2$ $57$ years $-11.8$ years $139.24$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{519.84} + {3.24} + {1310.44} + {139.24}} {{4}} $ $ {\sigma^2} = \dfrac{{1972.76}}{{4}} = {493.19\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{493.19\text{ years}^2}} = {22.2\text{ years}} $ The average turtle at the zoo is 68.8 years old. There is a standard deviation of 22.2 years.